# Discrete Normal Subgroup of Connected Group is Contained in Center

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## Theorem

Let $G$ be a connected topological group.

Let $\map Z G$ be the center of $G$.

Let $N$ be a discrete normal subgroup of $G$.

Then:

- $N \subseteq \map Z G$

## Proof

Let $h \in N$.

We shall show $h \in \map Z G$

Now, let:

- $A _h := \set {g \in G : g^{-1} h g = h}$

Then, we need to show:

- $A _h = G$

Let $e \in G$ denote the identity.

Since $e \in A _h$, we have $A _h \ne \O$.

Thus, it suffices to show that $A _h$ is clopen in view of Definition 4 of Connectedness.

As $N$ is a discrete subgroup, there is an open set $U \subseteq G$ such that:

- $U \cap N = \set e$

Multiplying the above equality from left, we also have:

- $\forall h \in N : h U \cap N = \set h$

Define $\phi _h : G \to N$ by:

- $\map {\phi _h} g := g^{-1} h g$

Then $\phi _h$ is continuous.

Thus both:

\(\ds A _h\) | \(=\) | \(\ds \set {g \in G : g^{-1} h g \in h U }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \phi _h^{-1} \sqbrk {h U}\) |

and

\(\ds G \setminus A _h\) | \(=\) | \(\ds \set {g \in G : g^{-1} h g \not \in h U }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \bigcup _{h' \in N \setminus \set h} \phi _h^{-1} \sqbrk {h' U}\) |

are open.

That is, $A _h$ is clopen.

This needs considerable tedious hard slog to complete it.In particular: More detailsTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |